Post on 26-Mar-2018
Four Questions• How many signals? Equivalence• Where on spectrum? Chemical Shift• How big? Integration• Shape? Splitting (coupling)
Basic Correlation Chart
• How many 1H signals?• Shift? Proton NMR Chemical Shift (ppm)
alkane CH3 0.9alkane CH2 1.3alkane CH 1.7X=C‐C‐H(Ph‐C‐H orO=C‐C‐H)
2‐3
O‐C‐HX‐C‐H
3‐4
(sp2) C=C‐H 5‐6Ph‐H 7‐8O=C‐H 9‐10O=C‐O‐H 10‐12O‐H, N‐H 1‐5
Four Questions• How many signals? Equivalence• Where on spectrum? Chemical Shift• How big? Integration• Shape? Splitting (coupling)
Integration
• Area under signal is proportional to the number of protons in the set
• RELATIVE AREA is calculated by computer
Integration
• Trace of spectrum is cut into sections above each signal to give integration values
Which proton set is this?
Integration
• You need to “set” the values• Here is a more realistic outcome
This is the NMR for a C5H10O compound. What are the integration values?
72.3 143.572.1
434.2
Product Distribution by Integration• In a mixture, areas are also proportional to number of protons
• But they can by fractionsA
B B
C
D E
F
G
F
DE
C
G
A
B F
57%43%
Four Questions• How many signals? Equivalence• Where on spectrum? Chemical Shift• How big? Integration• Shape? Splitting (coupling)
N + 1 Rule
• The signal for protons b is a quartet
• Signal b is “coupled” to the three protons labeled “a”
• Four possible affects on signal b• Summarized by this simplification: n+1, where n = number of adjacent, nonequivalent protons
Exchangeable protons
• If protons exchange, they are not coupled• If protons are not coupled, they will no show splitting in signal
• Typical for alcohols and acids
N+1 is a Simplification
• We learn it first because it applies to many molecules
• Assumes that all adjacent protons couple to an identical, measureable degree
• One exception: Aldehyde• We will learn a more robust treatment after we master first principles
Very small coupling between these protons, even though they are adjacent
O
H
H H
(of ten singlet)
(of ten doublet)
Predicting Spectra
• Step 2: Table of dataO
O
A B
CD
Signal Shift (ppm) Integration Splitting
A 2‐3 3 singlet
B 2‐3 2 triplet
C 3‐4 2 triplet
D 3‐4 3 singlet
Interpreting Spectra
• Reverse Process– Make table– Assign possible “pieces”– Predict structure
C6H10Cl2O2
2.3
4.5 4.6
4.6 7.0
• Make a table
C6H10Cl2O2
2.3
4.5 4.6
4.6 7.0
Peak Shift (ppm) Int Split
A 5.45 1 H T, n=2
B 4.1 2 H T, n=2
C 2.3 2 H Qt, n=3
D 2.1 2 H Qt, n=3
E 1.1 3 H T, n=2
• Assign “pieces”
C6H10Cl2O2
2.3
4.5 4.6
4.6 7.0
Peak Shift(ppm) Int Split Piece Adjacent
A 5.45 1 H T, n=2 CH Electronegative, CH2
B 4.1 2 H T, n=2 CH2 Oxygen? CH2
C 2.3 2 H Qt, n=3 CH2 CH3 or CH, CH2
D 2.1 2 H Qt, n=3 CH2 CH3 or CH, CH2
E 1.1 3 H T, n=2 CH3 CH2
‐Cl, ‐Cl, C=O
• Predict structure
C6H10Cl2O2
2.3
4.5 4.6
4.6 7.0
Peak Shift(ppm) Int Split Piece Adjacent
A 5.45 1 H T, n=2 CH Electronegative, CH2
B 4.1 2 H T, n=2 CH2 Oxygen? CH2
C 2.3 2 H Qt, n=3 CH2 CH3 or CH, CH2
D 2.1 2 H Qt, n=3 CH2 CH3 or CH, CH2
E 1.1 3 H T, n=2 CH3 CH2
‐Cl, ‐Cl, C=O
Review of Splitting
• Caused by shift due to magnetic fields of adjacent protons
• We say that these protons are “coupled”
• Protons may be coupled to different degrees• Coupling constant• Typically 7 Hz for adjacent sp3 carbons• Tree diagram
Basis of n+1 Rule• Shortcut: N+1 if all protons coupled with same constant
• Look at tree diagram
• Coupling constant is 7.1‐7.2 Hz
Tree diagram• Draw the tree diagram that shows why signal A is a triplet
• What is the coupling constant for signal A
Coupling Constants
• Coupling constants are not all 7Hz
• In this class, we will need to know other J values
Typical Constants
• Use the table to predict typical coupling constants
H3C
OCH3Ha
Hb
ab
ac
ad
bc
bd
cd
H
H3C CH3
H
ab
ac
ad
bc
bd
cd
Example: Cinnamic Acid
• Can the trans and cis isomers be differentiated using proton NMR?
• Yes—with coupling constants
7.40
7.33
7.40
7.60
7.60
11.0
7.45
6.33
OH
OH
H
7.40
7.33
7.40
7.60
7.60
11.0
7.35
5.98
OHO
H
H
400 MHz NMR:Doublet at 6.310ppm and6.355ppm
400 MHz NMR:Doublet at 5.925ppm and5.950ppm
Calculate the coupling constants for these doublets
Spectra that are not N+1
• Consider the allylic methyl group
• Coupled to two protons—but not with the same coupling constant
• Not N+1• Split into a doublet by Ha• That doublet is split into doublet by Hb
• It is doublet of doublets
Why not N+1?
• Jac = 1.7Hz (typical 0 Hz)
• Jbc = 6.9 Hz (typical 4‐10 Hz)
• If Jac = Jbc = 6.9Hz, what would we observe?
Predict the Splitting
• What signals would be observed for proton A?– Proton A is coupled to one proton B (doublet)
– Proton A is coupled to three proton C (quartet)
– Doublet of quartets with J = 15.6 Hz and J = 1.7Hz
Predict the Splitting
• Do the same for proton B
• How is signal same/different than proton A signal?
Exercise
• Explain these two observed signals– How would you describe them?
– Which proton(s) do they belong to?
Diastereotopic Protons
A
B C D
E
Peak multiplicity J (Hz)
A m
B sx 7.2
C dd 7.2, 14
D dd 7.2, 14
E d 7.2
Types of problems
• Know typical coupling constants• Describe expected signal (dd, dt, etc)• Draw expected proton NMR• Interpret proton NMR given coupling constants